- #1

Potatochip911

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- 3

## Homework Statement

Given two indistinguishable objects at the same initial temperature ##T_i##, calculate the minimum work done by a refrigerator functioning between the two objects till one of the objects reaches a new temperature ##T_2##, assume constant heat capacities and constant pressure.

a) Prove ##W^{min}_{on} = C_p[(\frac{T_i^2}{T_2})+T_2-2T_i]##

b) Now prove it using ##W_{by} \leq -\Delta A##, are there any other requirements necessary for the results from a) and b) to agree?

## Homework Equations

##dU = \delta Q_{in} + \delta W_{on}##

##\delta W_{quas} = -PdV##

##dS=\frac{\delta Q_{quas}}{T}##

##C_P=\left(\frac{\delta Q_{rev}}{dT}\right)_P##

## The Attempt at a Solution

[/B]

For part a:

Since it's constant pressure ##\delta Q_{rev} = C_pdT##, now since there's two objects I think the total heat is ##Q_{revT} = C_p\int_{T_i}^{T_2}dT + C_p\int_{T_i}^{T_h}dT## where ##T_h## is the final temperature of the object that isn't being cooled, I believe the heat of this object goes up. Factoring and integrating gives ##Q_{revT} = C_p(T_2-T_i+T_h-T_i) = C_p(T_2-2T_i+T_h)##

Edit: Ok I've managed to get the answer but frankly I don't know how to justify what I've done.

For the change in entropy we have $$\Delta S = C_p\int_{T_i}^{T_2}\frac{dT}{T}+C_p\int_{T_i}^{T_h}\frac{dT}{T} = C_p\ln(\frac{T_2}{T_i})+C_p\ln(\frac{T_h}{T_i})=C_p\ln(\frac{T_2T_h}{T_i^2})$$

Now if I set the change in entropy equal to zero I can get ##T_h = \frac{T_i^2}{T_2}## which plugging into the above expression gives the answer but I don't see why ##\delta S = 0## and it seems volume remains constant as well which gives the result ##W_{on} = -Q_{in}##

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